[quote]
…we are assuming an “infinite” number of 9s following the decimal point…**
But that criticism, though respectfully appreciated, bears the same flaw as with the classic inductive proof that my counter-proof counters. Such an assumption is a priori, and must be presented as axiomatic. One thing that has always bothered me about transfinitists is their inconsistent insistence that (1) infinity is not a number, but (2) there are an infinite number of so-and-so’s (9s, digits, whatever).
“It is lucky for rulers that men do not think.” — Adolf Hitler
Infinity is a number(s) (I’ll call the infinity we’re talking about omega). The problem with the argument:
1-.9999 is always greater than zero for a finite number of 9’s, therefore 1-.999… is greater than zero,
is flawed because induction shows that 1-.999… is greater than zero for any finite number of nines–induction does not make the step to showing that this is true for an infinite number of nines.
Consider the following argument:
1 is finite.
If n is finite, n+1 is also finite.
Therefore, the set of natural numbers is finite.
This seems to be equivalent to the flaw in the proof that .9999… is not 1–it works as long as we have finitely many 9’s, but not infinitely many 9’s. Induction doesn’t make that step to omega.
Of course it’s no different. (And as I understated, counting integer quantities, such as a baker’s dozen (13) in Base pi would remain “a bit harder.”) But as SqrlCub and Arnold Wrinkle-eyed later discussed, this was just the kind of thing SqrlCub was looking for. Useless (or, “usefulness unknown”), but easy to write.
The only thing I’d quibble is that, using the Arabic numerals and numbering convention with which we are all familiar, “pi” should be written 10, not 1. (Base ten, ten is written 10; Base six, six is written 10; etc. - and all deriving necessarily from the set of whole numbers.) If “pi” is 1 (the first, er…“whole” component of the series) AND also the base we use for describing larger numbers - then that’s Base One(?), in which you can’t calculate or even count anything. (I guess 1=>1, 2=>11, 3=>111…and that’s not a “counting” system.)
Dividing pi into a series to make Base pi countable is an arbitrary task (I vote for “bits,” in Base ten “1/8” pi), thus revealing the whole thing to be the nonsense (irrational sense that I can’t compute?) that it is.
So in order to disprove your hypothesis, all we must do is prove that no such thing exists.
We will do this by assuming it to be true and finding a contradiction (reductio ad absurdum).
Assume that the smallest positive non-zero real number eta exists.
Divide eta by two to obtain eta/2=zeta.
Zeta is not zero, because if it were, eta would also be zero (because eta=zeta*2).
But zeta is positive because eta is.
And furthermore, zeta is smaller than eta by the following proof:
zeta>0
zeta+zeta>0+zeta
eta/2+eta/2>zeta
eta>zeta.
So zeta is a smaller positive non-zero real number than eta.
But we assumed that eta was the smallest such number.
So the smallest positive non-zero real number must not exist.
I won’t get into the proof, but it can be proven that between any two arbitrarily close real numbers, there exists an infinite number of real numbers. Which is a consequence of the fact that there are more real numbers than integers (that is, the real numbers are not countable).
So, in conclusion, 0.999999… (with an infinite number of nines) is identical to 1 because the limit of 1-sum(9*10^(-n)) as n approaches infinity is 0.
This is not an assumption but a basic outgrowth of the definition of the real numbers. It is not an axiom and can be proven if you so desire, but it will involve me busting out my Real Analysis notes from college (without a doubt the hardest class I ever took in college or grad school).
So, it’s your belief that infinity is a number? In the top “flaw,” you claim that if there are n 9s after the decimal, there must be n-1 after multiplying by 10.
That can only be true if n is finite. It is not (that’s what the elipses mean in .999…). [infinity]/[any real number]=[infinity].
When I began to work on the question does .99999… equal 1, I believed that it didn’t but I wrote a proof that says it does. For the following proof, please realize that the sum of r^n from n=0 to infinity is 1/(1-r) where r < 1. Anyone who has studied Calculus will back me up on this. Also I don’t know if theres a way to type the summation symbol, so please bear with me.
Since we are dealing with a number with infinite decimal places, everyone should remember when dealing with infinity, what seems logical is not always correct. For instance the question of what is infinity divided by infinity?
Some would argue that anything divided by itself is 1. This is logical.
Others would argue that infinity divided by anything is still infinity. Logical.
Finally, others might argue that anything divided by infinity is 0. Also Logical.
What about infinity minus infinity?
Infinity is not a number, but an idea and does not follow the rules of numbers. Thats why functions are never considered at infinity, but rather as they approach infinity.
Heres another puzzler. The area under the curve of 1/(2^x) from 1 to infinity is actually finite.
This helps explain it better to me. My main problem with a number that has an infinite non-repeating series after it is that any product of that given multiplicant will also have an infinite series of numbers after it. I know what I was asking isn’t really practical but IMO I thought it would get rid of the infinite series in some way. Numbers are screwy that way.
Thanks for all the insight so far. I look forward to reading more in this topic.
HUGS!
Sqrl
Gasoline: As an accompaniement to cereal it made a refreshing change. Glen Baxter
Actually, just because one (or both) of the factors in a product has a nonrepeating decimal expansion doesn’t mean the product will. For example, both pi and 1/pi have nonrepeating decimal expansions, but they multiply to 1. There’s an infinite number of ways this can happen, and it’s not always easy to tell what kinds of arithmetic combinations of irrational/transcendental numbers will produce another irrational/transcendental.